Cho hàm số y=f(x) có đạo hàm trên tập

* Hướng dẫn giải

Đáp án A

Ta có $2f\left(x\right).\left[1-2x^2.f\left(x\right)\right]=x.f^{\text{'}}\left(x\right)\iff 2f\left(x\right)-4x^2.f^2\left(x\right)=x.f^{\text{'}}\left(x\right)$

$\iff 2x.f\left(x\right)-4x^3.f^2\left(x\right)=x^2.f^{\text{'}}\left(x\right)$

$\iff \frac{2x.f\left(x\right)-x^2.f^{\text{'}}\left(x\right)}{f^2\left(x\right)}=4x^3$

$\iff {\left(\frac{x^2}{f\left(x\right)}\right)}^{\text{'}}=4x^3\iff \int {\left(\frac{x^2}{f\left(x\right)}\right)}^{\text{'}}dx=\int 4x^{3}dx\iff \frac{x^2}{f\left(x\right)}=x^4+C\iff f\left(x\right)=\frac{x^2}{x^4+C}$.

Mà $f\left(2\right)=\frac{2}{3}\iff \frac{2^2}{2^4+C}=\frac{2}{3}\iff C=-10\Rightarrow f\left(x\right)=\frac{x^2}{x^4-10}$.

Ta có $\underset{1}{\overset{3}{\int }}f\left(x\right).\left(x^3-\frac{10}{x}\right)dx=\underset{1}{\overset{3}{\int }}\frac{x^2}{x^4-10}.\left(x^3-\frac{10}{x}\right)dx=4$.